Three forms of energy possessed by a flowing fluid are as follows:
1. Pressure energy:
Let an ideal fluid of density P be contained in a rectangular vessel, provided with a small side tube at a depth h_o below the free surface of fluid in the vessel. At the level of side tube, pressure of fluid along the axis of side tube
p = h₀pg
If we want to introduce more fluid into the vessel at this very pressure, we can force it through the side tube by doing work on the piston, If 'A' be the cross-section area of the piston, then force acting on the piston F = PA.
\(\therefore\) Work done in moving the piston through a small distance \(\Delta r\) will be
\(\Delta W=F\Delta x=PA\Delta x=P\Delta V\)
As a result of motion of piston, the mass of the fluid forced in the vessel
\(\Delta m=\rho A\Delta x=\rho\Delta V\)
The work done is stored up in the liquid in the form of its pressure energy.
\(\therefore\) Pressure energy of liquid per unit mass = Work done per unit mass
 \(=\frac{\Delta W}{\Delta m}=\frac{P\Delta V}{\rho\Delta V}=\frac{p}{\rho}\)
and pressure energy per unit volume = p.
2. Gravitational potential energy:
Let at any stage of its flow a fluid element of mass "m' be situated at a height 'h' from the reference line (generally taken to be earth's surface), then its gravitational potential energy in that position is mgh.
\(\therefore\) Gravitational potential energy per unit mass = \(\frac{mgh}{m}=gh\)
and gravitational potential energy per unit volume = pgh.
3. Kinetic energy:
Let at any stage of its flow, a fluid element of mass 'm' be moving with a speed 'v', then the kinetic energy of this fluid element is \(\frac{1}{2}v^2\)
\(\therefore\) Kinetic energy per unit mass = \(\frac{\frac{1}{2}mv^2}{m}=\frac{1}{2}v^2\)
and kinetic energy per unit volume = \(\frac{1}{2}\rho v^2\)
These three forms of energy possessed by a flowing fluid are mutually convertible from one form to another.

(a) Inside a liquid drop
Let r = radius of a spherieal liquid drop of centre O. T = surface tension of the liquid. Let P_i and P_o be the values of pressure inside and outside the drop.
\(\therefore\) Excess pressure inside the liquid drop = P_i - P_o
Let \(\Delta\)r be the increase in its radius due to excess pressure. It has one free surface outside.
\(\therefore\)  increase in surface area of the liquid drop
= \(4\pi(r+\Delta r)^2-4\pi r^2\)
= \(4\pi[r^2+(\Delta r)^2+2r\Delta r-r^2]\)
= \(8\pi r\quad \Delta r\)  .................(i)
\(\therefore\) increase in surface energy of the drop is
W = Surface tension x increase in area 
= \(T\times 8\pi r\quad \Delta r\)  .............. (ii)
Also W = Force due to excess of pressure x displacement
= Excess pressure x Area of drop x increase in radius
= \((p_i-p_0)4\pi r^2\Delta r\) ............(iii)
\(\therefore\) From eqns (ii) and (iii), we get
\((p_i-p_0)\times 4\pi r^2\quad \Delta r=T\times 8\pi r\quad \Delta r\)
\(\Rightarrow p_i-p_0=\frac{2T}{r}\)
(b) Inside a liquid bubbles:
A liquid bubble has air both inside and outside it and therefore it has two free surfaces.
Thus increase in its surfaces area
= \(2[4\pi (r+\Delta r)^2-4\pi r^2]\)
= \(2\times 8\pi r\Delta r=16\pi r\quad\Delta r\)
\(\therefore\)   W = \(T\times 16\pi r\Delta r\)     ...............(i)
Also  W = \((p_i-p_0)4\pi r^2\times \Delta r\)  ........(ii)
From eqnd (i) and (ii), we get
\((p_i-p_0)\times 4\pi r^2\times \Delta r=T.16\pi r\quad \Delta r\)
or \((p_i-p_0)=\frac{4T}{r}\)
(c) Inside an air bubble:
Air bubble is formed inside liquid, thus air bubble has one free surface inside it and liquid is outside.
If   r = radius of air bubble
\(\Delta\)r = increase in its radius due to excess of pressure
(P_i - P₀) inside it.
T = surface tension of the liquid in which bubble is formed.
\(\therefore\) increase in surface area = \(8\pi r \Delta r\).
\(\therefore\)    W = T x \(8\pi r \Delta r\)
Also W = \((p_i-p_0)\times 4\pi r^2\Delta r\)
\(\therefore (p_i-p_0)\times 4\pi r^2\Delta r=T\times 8\pi r^2\Delta r\)
or  \(p_i-p_0=\frac{2T}{r}\)

The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m. Thus, the shear force F is
F = T = mg = 0.010 x 9.8 = 9.8 x 10^-2 N
Shear stress on the fluid
\(\frac { F }{ A } =\frac { 9.8\times { 10 }^{ -2 } }{ 0.10 } =0.98N/{ m }^{ 2 }\)
Strain rate,
\(\eta =\frac { stress }{ Strain\ rate } s^{-1}\) 
\( =\frac{\left(9.8 \times 10^{-2} \mathrm{~N}\right)\left(0.30 \times 10^{-3} \mathrm{~m}\right)}{\left(0.085 \mathrm{~m} \mathrm{~s}^{-1}\right)\left(0.10 \mathrm{~m}^2\right)}\) 
\( =3.46 \times 10^{-3} \mathrm{~Pa} \mathrm{~s} \)

Since pressure is transmitted undiminished throughout the fluid,
\(F_{1}=\frac{A_{1}}{A_{2}} F_{2}=\frac{\pi\left(5 \times 10^{-2} \mathrm{~m}\right)^{2}}{\pi\left(15 \times 10^{-2} \mathrm{~m}\right)^{2}}\left(1350 \mathrm{~kg} \times 9.8 \mathrm{~ms}^{-2}\right)\)
= 1470 N
≈ 1.5 x 10³ N
The air pressure that will produce this force is
\(P=\frac{F_{1}}{A_{1}}=\frac{1.5 \times 10^{3} \mathrm{~N}}{\pi\left(5 \times 10^{-2}\right)^{2} \mathrm{~m}}=1.9 \times 10^{5} \mathrm{~Pa}\)
This is almost double the atmospheric pressure.

Because terminal velocity of air bubble is negative.

For streamline or laminar flow 0 < NR < 2000,
for turbulent flow, NR > 3000
If N_R lies between 2000 to 3000, the flow is unstable i.e., change from streamline flow to turbulent flow.

(i) \(\frac{v^2}{2g}\)
(ii) \(\frac{p}{\rho g}\)

Given, mass of girl (m) = 50kg
Diameter of circular heel (2r) = 1.0cm
\(\therefore \)Radius (r) = 0.5 cm = \(5\times 10^{ -3 }\)m
Area of circular heel (A) = \({ \pi r }^{ 2 }=3.14\times (5\times 10^{ -3 })^{ 2 }m^{ 2 }\)
\(=78.50\times 10^{ -6 }m^{ 2 }\)
\(\therefore \) Pressure exerted on the horizontal floor
\(\rho =\frac { F }{ A } =\frac { mg }{ A } =\frac { 50\times 9.8 }{ 78.50\times 10^{ -6 } } =6.24\times 10^{ 6 }Pa\)

The difference between absolute pressure and atmospheric p[ressure is known as gauge pressure.
As,p_absolute= p_a+ρgh
So,p_absolute− p_a=ρghi.e.p_gauge=ρgh
Here,ρ is the density of a fluid of depth h.

The critical speed of a fluid depends on ( a ) diameter of tube,  ( b ) density of fluid, ( c ) coefficient of viscosity of the fluid.

This is due to equation of continuity, a₁ v_I = a₂ v₂  
∵∵  a₁  > a₂
∴∴     v₂  > v₁

\(Given,\ radius\ (r)=2.5\times { 10 }^{ -5 }m\)
\(\\ Surface\ tension(S)=7.28\times { 10 }^{ -2 }N/m\)
\(\\ Angle\ of\ contact\ (\theta )={ 0 }^{ 0 }\)
\(\\ density\ of\ water\ (\rho )={ 10 }^{ 3 }kg/{ m }^{ 3 }\)
\(\\ The\ maximum\ height\ which\ sap\ can\ rise\ in\ trees\ through\ capillarity\ action\ given\ by\)
\(\\ h=\frac { 2Scos\theta }{ r\rho g } \)
\(\\ h=\frac { 2\times 7.28\times { 10 }^{ -2 }\times cos\theta }{ 2.5\times { 10 }^{ -5 }\times 1\times { 10 }^{ 3 }\times 9.8 } \)
But, the height of many trees are more than 0.59m, therefore, the rise of sap in all trees is not possible through capillarity action alone.

Soap bubble bursts after some time because the pressure inside it become more than the outside pressure.

Radius, r = 1 cm; p = 0.9 g cm^-3
h = 1.36 mm = 0.136 cm
Pressure, p = hpg = 0.136 x 0.9 x 980
= 119.95 dyne cm^-2
Let T be surface tension of soap solution
\(\therefore\) Excess pressure, p = \(\frac{4T}{r}\)
or  T = \(\frac{Pr}{4}\)
= \(\frac{119.95\times 1}{4}\)
= 29.988 dyne cm^-1.

Height of water column, h₁ = 10.0 cm
Density of water, \({ \rho }_{ 1 }\) = 1g/ cm 3
Height of spirit column, h₂ = 12.S cm
Density of spirit,\({ \rho }_{ 2 }\) =?
The mercury column in both arms of the U-tube are at same level, therefore, pressure in both arms will be same.
Pressure exerted by water column = pressure exerted by spirit column
p₁= p₂
\({ h }_{ 1 }{ \rho }_{ 1 }g={ h }_{ 2 }{ \rho }_{ 2 }g\ or\ { \rho }_{ 2 }=\frac { { h }_{ 1 }{ \rho }_{ 1 } }{ { h }_{ 2 } } =\frac { 10\times 1 }{ 12.5 } =0.80g/{ cm }^{ 3 }\) 
Specific gravity of spirit=\(\frac { Density\ of\ spirit }{ Density\ of\ water } =\frac { 0.80 }{ 1 } =0.80\)